Answer:
[tex]\large\boxed{\{-7,\ 1\}}[/tex]
Step-by-step explanation:
[tex]x^2+6x=7\\\\x^2+2(x)(3)=7\qquad\text{add}\ 3^2=9\ \text{to both sides}\\\\x^2+2(x)(3)+3^2=7+9\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+3)^2=16\Rightarrow x+3=\pm\sqrt{16}\\\\x+3=-4\ \vee\ x+3=4\qquad\text{subtract 3 from both sides}\\\\x=-7\ \vee\ x=1[/tex]
